Let $h$ be a function defined for all real numbers except for $0$. Also let $h'$, the derivative of $h$, be defined as $h'(x)=\dfrac{(x+3)}{x^2}$. On which intervals is $h$ decreasing? Choose 1 answer: Choose 1 answer: (Choice A) A $x>0$ and $-3<x<0$ (Choice B) B $x<-3$ only (Choice C) C $-3<x<0$ only (Choice D) D $x<-3$ and $-3<x<0$ (Choice E) E The entire domain of $h$
Explanation: We can analyze the intervals where $h$ is increasing/decreasing by looking for the intervals where its derivative $h'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $h'(x)=\dfrac{(x+3)}{x^2}$ and that $h$ is undefined at $x=0$. $h'(x)=0$ for $x=-3$. Our critical point is $x=-3$, and we should also consider $x=0$. Our points divide the number line into three intervals: $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $\begin{array}{rl} x<& \llap{-}3\end{array}$ $ \llap{-}3<x<0$ $ x>0$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $x<-3$ $x=-4$ $h'(-4)=-\dfrac{1}{16}<0$ $h$ is decreasing $\searrow$ $-3<x<0$ $x=-1$ $h'(-1)=\dfrac14>0$ $h$ is increasing $\nearrow$ $x>0$ $x=1$ $h'(1)=4>0$ $h$ is increasing $\nearrow$ In conclusion, $h$ is decreasing over the interval $x<-3$ only.